Dear Laurent (and Gianni), So if the following code does useless work:
perl6 -pe '.chop' demo1.txt why doesn't it fail with an error, "Useless use of ... in sink context (line 1)"? Best, Bill. On Tue, May 5, 2020 at 12:43 PM Laurent Rosenfeld <laurent.rosenf...@googlemail.com> wrote: > > In: > perl6 -ne 'put .chop' demo1.txt > > the script prints out the value returned by the chop method, because put acts > on this value. > > In: > perl6 -pe '.chop' demo1.txt > the value returned by chop is discarded and the script print $_ unaltered. > > Cheers, > Laurent. > > Le mar. 5 mai 2020 à 21:07, William Michels via perl6-users > <perl6-us...@perl.org> a écrit : >> >> On Tue, May 5, 2020 at 8:01 AM Gianni Ceccarelli <dak...@thenautilus.net> >> wrote: >> > >> > On 2020-05-05 William Michels via perl6-users <perl6-us...@perl.org> >> > wrote: >> > > mbook:~ homedir$ perl6 -ne 'put .chop' demo1.txt >> > > this is a test >> > > I love Unix >> > > I like Linux too >> > > mbook:~ homedir$ perl6 -pe '.chop' demo1.txt >> > > this is a test, >> > > I love Unix, >> > > I like Linux too, >> > >> > The ``.chop`` method does not mutate its argument, it only returns a >> > chopped value. If you want to mutate, you need to say so:: >> > >> > raku -pe '.=chop' demo1.txt >> > >> > Notice that the ``.=`` operator is: >> > >> > * not specific to ``chop`` >> > * not even specific to calling methods >> > >> > In the same way that ``$a += 1`` is the same as ``$a = $a + 1``, so >> > ``$a .= chop`` is the same as ``$a = $a.chop``. >> > >> > So, if you wanted, you could do:: >> > >> > raku -pe '.=uc' # print upper-case >> > >> > -- >> >> I appreciate the reply, but your answer fails to explain one thing: >> why does chop work without ".=" assignment using the "-ne" one-liner >> flag, but not with the "-ne" one-liner flag"? According to the help >> screen (running 'perl6 -help' at the bash command prompt), this is >> what it says about the "-n" and the "-e" flags: >> >> -n run program once for each line of input >> -p same as -n, but also print $_ at the end of lines >> >> So what strikes me from the definitions above is the part where "-p" >> is the "same as -n... (with autoprinting of $_)." That leads people to >> believe that they can write a short one-liner with the -ne flag ('put >> .chop') and an even shorter one-liner with the -pe flag ('.chop'). >> >> If the only difference between the "-n" and "-p" flags is really that >> the second one autoprints $_, I would have expected the "-pe" code >> above to work identically to the "-ne" case (except "-ne" requires a >> print, put or say). Presumably 'perl6 -ne "put .chop" ' is the same >> as 'perl6 -ne ".chop.put" ' , so if ".put" isn't returning $_ , >> what is it returning then? >> >> Look, It's no big deal if I have to write 'perl6 -pe ".=chop" ' >> instead of 'perl6 -pe ".chop" ', I just want to resolve in my mind a >> perceived inconsistency wherein there's no requirement to write >> 'perl6 -ne "put .=chop" ' for the "-ne" case, but there IS a >> requirement to write 'perl6 -pe ".=chop" ' for the "-pe" case. >> >> Best Regards, Bill.